Fix the Code: Un-Mutating an Array

Your friend is trying to write a function to accomplish the following transformations:

let x = [3, 3, 3, 3, 3, 3, 3]

// Each time x is called, the following results are shown:

change(x, 0)  // [3, 3, 3, 3, 3, 3, 3]
change(x, 1)  // [3, 2, 2, 2, 2, 2, 3]
change(x, 2)  // [3, 2, 1, 1, 1, 2, 3]
change(x, 3)  // [3, 2, 1, 0, 1, 2, 3]

Note: The change() function should not mutate the original array. After each call to the function, the original x should still equal [3, 3, 3, 3, 3, 3, 3].

He comes up with the following code:

function change(x, times) {
  for(let i = 0; i < x.length; i++) {
    let j = 1;
    while (j <= times) {
      if (i >= j && i < x.length-j) {
        x[i]--;
      }
      j++;
    }
  }
  return x;
}

Oops! The code appears to mutate the original array. Fix this incorrect code so that the function no longer mutates the original array.

See below:

Examples

let x = [3, 3, 3, 3, 3, 3, 3]

// What we want:
change(x, 2) => [3, 2, 1, 1, 1, 2, 3]

change(x, 2) => [3, 2, 1, 1, 1, 2, 3]

// What we get:
change(x, 2) => [3, 2, 1, 1, 1, 2, 3]  // Good so far...

change(x, 2) => [3, 1, -1, -1, -1, 1, 3] // Array is mutated :(

Notes

• If this is confusing, copy and paste the incorrect code in a REPL environment and play around with the code to understand what the function is doing.
• Hint: Try to make a copy of the input array.
• If this looks familiar, it is part of a solution for the Concentric Rug problem.